Problem:
Solution:
This is a really bored after lunch hour, so I decided let's do a simple calculus exercise to wake my brain up.
The 11+u2 reminded me this has something to do with tan. So let's do this simple integration.
∫du1+u2=∫sec2θdθ1+tan2θ=∫dθ=θ=tan−1u.
That's refresh my memory! 11+u2 is the derivative of tan−1u.
Now we apply the Lagrange's mean value theorem to get:
tan−1v−tan−1uv−u=11+δ2.
Where u<δ<v.
tan−1v−tan−1u=v−u1+δ2.
Therefore we conclude (remember when you increase the denominator, the value decrease, and vice versa):
v−u1+v2<tan−1v−tan−1u<v−u1+u2.
Last but not least, set u=1 and v=43 gives
43−11+(43)2<tan−143−tan−11<43−11+12.
325<tan−143−tan−11<16.
π4+325<tan−143<π4+16.
Solution:
This is a really bored after lunch hour, so I decided let's do a simple calculus exercise to wake my brain up.
The 11+u2 reminded me this has something to do with tan. So let's do this simple integration.
∫du1+u2=∫sec2θdθ1+tan2θ=∫dθ=θ=tan−1u.
That's refresh my memory! 11+u2 is the derivative of tan−1u.
Now we apply the Lagrange's mean value theorem to get:
tan−1v−tan−1uv−u=11+δ2.
Where u<δ<v.
tan−1v−tan−1u=v−u1+δ2.
Therefore we conclude (remember when you increase the denominator, the value decrease, and vice versa):
v−u1+v2<tan−1v−tan−1u<v−u1+u2.
Last but not least, set u=1 and v=43 gives
43−11+(43)2<tan−143−tan−11<43−11+12.
325<tan−143−tan−11<16.
π4+325<tan−143<π4+16.
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